An Application of Number Theory, the RSA Cryptosystemfaculty.washington.edu/moishe/hanoiex/Number...

An Application of Number Theory, the RSACryptosystem

Ngày 10 tháng 12 năm 2010

An Application of Number Theory, the RSA Cryptosystem

Securing Transactions

QuestionMr. Nguyen sells expensive jewelry. He has an interesting ideafor a business model. Each customer will have access to boxeswith a combination lock. Once a person grabs a box he can sethis own private combination lock. An open box can be closedby anyone, but only the owner knows the combination and canopen it. The content of any open box sent between persons willbe stolen.You wish to buy an expensive gift for your significant other’sbirthday. This means money will have to be sent to Mr. Nguyen(who is honest and trustworthy) and the gift delivered to you.Transaction details, such as item, price etc. can be discussedby phone.

How can we accomplish this?

DiscussionThis is exactly how business transactions are being conductedon the internet today, except that the boxes are virtual boxes.Closing a box is accomplished by encrypting the message. Sowhile the message is traveling on the internet, being exposed tohackers and others, it is encrypted using a “key”. Only theowner of the key knows how to open the box and retrieve itscontent.

QuestionThe question faced scientists was how to design a system withthe following properties:

1 A group of particpants can securely communicate witheach other over an open system.

2 How can anyone send a message to bob so no one exceptBob will be able to understand the message.

3 Can messages be “signed’?.

RSA Public Key System

DiscussionUntil the mid-70’s encryptions were done using private keys.Two persons or institutions that needed to establish securecommunications shared a private key they used for encryption.

The system worked quite well, except for one problem: how toshare keys.

DES, (Data Encryption Standard) was a popular private keysystem that was widely used by many governments andinstitutions.

It was recently replaced by another system, AES (AdvancedEncryption Standard).

In 1976 Rivest, Shamir and Adelman proposed the public keycryptosystem: RSA.

It was recently replaced by another system, AES (AdvancedEncryption Standard).

In 1976 Rivest, Shamir and Adelman proposed the public keycryptosystem: RSA.

It was recently replaced by another system, AES (AdvancedEncryption Standard).In 1976 Rivest, Shamir and Adelman proposed the public keycryptosystem: RSA.

Each key consisted of two parts, a public part used forencryption and a private part used for decryption.

1 Every message can be coded as an integer M .

2 Public key: (K ,e) where K = pq, p,q prime numbers,gcd(e, (p − 1)(q − 1)) = 1.

3 To encrypt the message M , coded as an integer, wecalculate EM = Me mod K and send EM to the owner ofthe public key (K ,e).

4 Decryption: The key owner first findsd = e−1 mod (p − 1)(q − 1).

5 To retrieve M , the owner of the key (K ,e) calculates:EMd mod K = M .

1 Every message can be coded as an integer M .2 Public key: (K ,e) where K = pq, p,q prime numbers,

gcd(e, (p − 1)(q − 1)) = 1.

3 To encrypt the message M , coded as an integer, wecalculate EM = Me mod K and send EM to the owner ofthe public key (K ,e).

gcd(e, (p − 1)(q − 1)) = 1.3 To encrypt the message M , coded as an integer, we

calculate EM = Me mod K and send EM to the owner ofthe public key (K ,e).

Prime numbers, key facts

The following theorems play a central role discussing primesand factorization.

1 T1: GF (p) = {0,1, . . . ,p − 1} is a field (addition andmultiplication are done mod p)

2 T2: GF (p) has primitve elements. α ∈ GF (p) is primitive if{αi | i = 0,1, . . .p − 2} = {0,1,2, . . . ,p − 1}.

3 T3: If p(x) is a polynomial with coefficients in GF (p) andf (β) = 0 then p(x) = (x − β)p1(x) where p1(x) is apolynomial with coefficients in GF (p).

4 T4: A finite field has pn (p prime) elements and is uniqueupto isomorphism.

Theorem (Fermat’s theorem)

If p is prime and a < p then ap−1 mod p = 1.

Chứng minh.

Since GF (p) is a field for anya ∈ GF (p) {a,2a,3a, . . . , (p − 1)a} = {1,2,3, . . . ,p − 1}.So a · 2a · 3a · · · (p − 1)a = 1 · 2 · 3 · · · (p − 1)ap−1 ·

∏p−1i=1 i =

∏p−1i=1 i mod p ⇒ ap−1 = 1 mod p

Theorem (Fermat’s theorem)

If p is prime and a < p then ap−1 mod p = 1.

Chứng minh.

Since GF (p) is a field for anya ∈ GF (p) {a,2a,3a, . . . , (p − 1)a} = {1,2,3, . . . ,p − 1}.So a · 2a · 3a · · · (p − 1)a = 1 · 2 · 3 · · · (p − 1)ap−1 ·

∏p−1i=1 i =

∏p−1i=1 i mod p ⇒ ap−1 = 1 mod p

Fermat’s theorem can be applied to simplify some intimidatinglooking computations:

Example

1 Calculate 7341235 mod 11.

2 11 is prime, so 710 mod 11 = 17341235 mod 11 = (710)34123 · 75 mod 11 = 75 mod 11 = 10

4 341 = 31 · 11, 7341235 mod 11 = 10.7341235 mod 31 = 7341220715 mod 31 = 715 mod 31.72 mod 31 = 18, 73 mod 31 = 7 · 18 mod 31 =126 mod 31 = 275 mod 31 = 73 · 72 mod 31 = 1.So if x = 7341235 then we have:x mod 11 = 10, x mod 31 = 1.We can now use the Chinese Reaminder Theorem and get:7341235 mod 341 = 32.

Example1 Calculate 7341235 mod 11.

Example1 Calculate 7341235 mod 11.2 11 is prime, so 710 mod 11 = 1

7341235 mod 11 = (710)34123 · 75 mod 11 = 75 mod 11 = 103 Calculate 7341235 mod 341.

7341235 mod 11 = (710)34123 · 75 mod 11 = 75 mod 11 = 10

7341235 mod 11 = (710)34123 · 75 mod 11 = 75 mod 11 = 103 Calculate 7341235 mod 341.4 341 = 31 · 11, 7341235 mod 11 = 10.

7341235 mod 31 = 7341220715 mod 31 = 715 mod 31.72 mod 31 = 18, 73 mod 31 = 7 · 18 mod 31 =126 mod 31 = 275 mod 31 = 73 · 72 mod 31 = 1.So if x = 7341235 then we have:x mod 11 = 10, x mod 31 = 1.We can now use the Chinese Reaminder Theorem and get:7341235 mod 341 = 32.

7341235 mod 31 = 7341220715 mod 31 = 715 mod 31.

72 mod 31 = 18, 73 mod 31 = 7 · 18 mod 31 =126 mod 31 = 275 mod 31 = 73 · 72 mod 31 = 1.So if x = 7341235 then we have:x mod 11 = 10, x mod 31 = 1.We can now use the Chinese Reaminder Theorem and get:7341235 mod 341 = 32.

7341235 mod 31 = 7341220715 mod 31 = 715 mod 31.72 mod 31 = 18, 73 mod 31 = 7 · 18 mod 31 =126 mod 31 = 2

75 mod 31 = 73 · 72 mod 31 = 1.So if x = 7341235 then we have:x mod 11 = 10, x mod 31 = 1.We can now use the Chinese Reaminder Theorem and get:7341235 mod 341 = 32.

7341235 mod 31 = 7341220715 mod 31 = 715 mod 31.72 mod 31 = 18, 73 mod 31 = 7 · 18 mod 31 =126 mod 31 = 275 mod 31 = 73 · 72 mod 31 = 1.

So if x = 7341235 then we have:x mod 11 = 10, x mod 31 = 1.We can now use the Chinese Reaminder Theorem and get:7341235 mod 341 = 32.

7341235 mod 31 = 7341220715 mod 31 = 715 mod 31.72 mod 31 = 18, 73 mod 31 = 7 · 18 mod 31 =126 mod 31 = 275 mod 31 = 73 · 72 mod 31 = 1.So if x = 7341235 then we have:x mod 11 = 10, x mod 31 = 1.

We can now use the Chinese Reaminder Theorem and get:7341235 mod 341 = 32.

7341235 mod 31 = 7341220715 mod 31 = 715 mod 31.72 mod 31 = 18, 73 mod 31 = 7 · 18 mod 31 =126 mod 31 = 275 mod 31 = 73 · 72 mod 31 = 1.So if x = 7341235 then we have:x mod 11 = 10, x mod 31 = 1.We can now use the Chinese Reaminder Theorem and get:7341235 mod 341 = 32.

Fermat’s theorem can be used to test whether an integer iscomposite (not prime).

Given an integer n, if an−1 mod n 6= 1, a < n then n iscomposite.

But what if an−1 = 1?For example: 2340 mod 341 = 1 but 341 = 11 · 31a1728 mod 1729 = 1 ∀a relatively prime to 1729.

QuestionCan you prove it? It is not difficult, give it a try.

Question (Challenge)The other day we found the 163 digits long key below on theinternet. It is not prime, easy to check.2Key−1 mod Key 6= 1Can we find its prime factors?

Key =1193098423264097759646037965385887599016380476452728541299175513582355781793126309459269365733778030509749311859187902804005784261377727067235425553086083970158319

QuestionAre there any other ways to factor integers besides trying theGCD of the integer with smaller integers?

Question (Challenge)The other day we found the 163 digits long key below on theinternet. It is not prime, easy to check.2Key−1 mod Key 6= 1Can we find its prime factors?

Key =1193098423264097759646037965385887599016380476452728541299175513582355781793126309459269365733778030509749311859187902804005784261377727067235425553086083970158319

QuestionAre there any other ways to factor integers besides trying theGCD of the integer with smaller integers?

Is p really a prime number?

QuestionIs there a way to certify that a given number p is indeed prime?

Theorem (Wallis)

p is prime if and only if (p − 1)! mod p = −1.

Is p really a prime number?

QuestionIs there a way to certify that a given number p is indeed prime?

Theorem (Wallis)

p is prime if and only if (p − 1)! mod p = −1.

Miller-Rabin Test

Let N be an integer. By Fermat’s theorem if N is prime thenaN−1 mod N = 1. This calculation can be executed very fast onintegers with a few thousand digits. This means that if for some1 < a < N − 1, aN−1 mod N 6= 1 then N is definitely not aprime number.

But what can we conclude if aN−1 mod N = 1?

Answer: NOTHING. N can be composite, or prime.

Example

1 2340 mod 341 = 1 but 341 = 11 · 31.2 But 3340 mod 341 = 56 proves that 341 is composite.3 On the other hand, if

gcd(a,1729) = 1 then a1728mod1729 = 1.4 Since φ(1729) = 1729(1− 1

7)(1−1

13)(1−1

19) = 1296 if weselect a randomly we do not have a good chance to find aninteger that will prove that 1729 is not a prime number .

Miller-Rabin Test

Example

7)(1−1

13)(1−1

Miller-Rabin Test

Example

7)(1−1

13)(1−1

Miller-Rabin Test

Example1 2340 mod 341 = 1 but 341 = 11 · 31.

2 But 3340 mod 341 = 56 proves that 341 is composite.3 On the other hand, if

7)(1−1

13)(1−1

Miller-Rabin Test

Example1 2340 mod 341 = 1 but 341 = 11 · 31.2 But 3340 mod 341 = 56 proves that 341 is composite.

3 On the other hand, ifgcd(a,1729) = 1 then a1728mod1729 = 1.

4 Since φ(1729) = 1729(1− 17)(1−

113)(1−

119) = 1296 if we

select a randomly we do not have a good chance to find aninteger that will prove that 1729 is not a prime number .

Miller-Rabin Test

Example1 2340 mod 341 = 1 but 341 = 11 · 31.2 But 3340 mod 341 = 56 proves that 341 is composite.3 On the other hand, if

gcd(a,1729) = 1 then a1728mod1729 = 1.

4 Since φ(1729) = 1729(1− 17)(1−

113)(1−

119) = 1296 if we

select a randomly we do not have a good chance to find aninteger that will prove that 1729 is not a prime number .

Miller-Rabin Test

Example1 2340 mod 341 = 1 but 341 = 11 · 31.2 But 3340 mod 341 = 56 proves that 341 is composite.3 On the other hand, if

7)(1−1

13)(1−1

Miller-Rabin Test

Numbers N like 1729 for which aN−1 mod N = 1 ∀a relativelyprime to N are called Carmichael numbers. They are rare, butthere are infinitely many of them.

So Fermat’s theorem is not a good test for primality. We need abetter test.

Theorem (Miller-Rabin Test)

Let N be an integer, N − 1 = 2m · (2k + 1).An integer n is NOT a “composite-witness” for N if:

1 For some 1 ≤ i ≤ m, n(2k+1)2imod N = −1.

2 Or n(2k+1)2imod N = 1 and n2k+1 mod N = 1

In other words, the test fails to prove that N is composite.

Miller-Rabin Test

Chứng minh.

If p is prime then by Fermat’s theorem ap−1 mod p = 1.So a(p−1)/2 mod p =

√1 = ±1.

If a(p−1)/2 mod p = −1 then the test stops. In other words, it willnot say that p is composite.If a(p−1)/2 mod p = 1 then we calculate a(p−1)/4 mod p = ±1We continue until we reach a2k+1 mod p

We skip the important part of the proof. They proved that if N iscomposite then more than 50% of the integers a < N will becomposite-witnesses. In other words, to test whether an integerp is prime, we randomly select say 100 integers a < p andapply to them the Miller-Rabin test. If the test fails, we assumethat p is a prime number. The probabilty that we made amistake, that is decided that p is prime while in fact it is not, isless then

Example

1 1729 is a composite integer. Indeed 1728 = 26 · 33 and31728/2i

mod 1729 = 1 but 31728/64 mod 1729 = 664proving that 1729 is composite.

2 c = 9746347772161 is a Carmichael number.3 39746347772160 mod 9746347772161 = 1.4 39746347772160/2 mod 9746347772161 = 1, no decission.

So we continue.5 39746347772160/4 mod 9746347772161 = 4485448662696

proving that c is composite.

Example1 1729 is a composite integer. Indeed 1728 = 26 · 33 and

31728/2imod 1729 = 1 but 31728/64 mod 1729 = 664

proving that 1729 is composite.

2 c = 9746347772161 is a Carmichael number.3 39746347772160 mod 9746347772161 = 1.4 39746347772160/2 mod 9746347772161 = 1, no decission.

31728/2imod 1729 = 1 but 31728/64 mod 1729 = 664

proving that 1729 is composite.2 c = 9746347772161 is a Carmichael number.

3 39746347772160 mod 9746347772161 = 1.4 39746347772160/2 mod 9746347772161 = 1, no decission.

31728/2imod 1729 = 1 but 31728/64 mod 1729 = 664

proving that 1729 is composite.2 c = 9746347772161 is a Carmichael number.3 39746347772160 mod 9746347772161 = 1.

4 39746347772160/2 mod 9746347772161 = 1, no decission.So we continue.

5 39746347772160/4 mod 9746347772161 = 4485448662696proving that c is composite.

31728/2imod 1729 = 1 but 31728/64 mod 1729 = 664

proving that 1729 is composite.2 c = 9746347772161 is a Carmichael number.3 39746347772160 mod 9746347772161 = 1.4 39746347772160/2 mod 9746347772161 = 1, no decission.

So we continue.

5 39746347772160/4 mod 9746347772161 = 4485448662696proving that c is composite.

31728/2imod 1729 = 1 but 31728/64 mod 1729 = 664

proving that 1729 is composite.2 c = 9746347772161 is a Carmichael number.3 39746347772160 mod 9746347772161 = 1.4 39746347772160/2 mod 9746347772161 = 1, no decission.

Factoring

Discussion

1 To implement RSA we need to manufacture large primes.2 The Miller-Rabin test is commonly used for this purpose.3 There are also efficient algorithms to manufacture

“certified” primes.4 Are all large primes safe?

Factoring

Discussion1 To implement RSA we need to manufacture large primes.

2 The Miller-Rabin test is commonly used for this purpose.3 There are also efficient algorithms to manufacture

Factoring

Discussion1 To implement RSA we need to manufacture large primes.2 The Miller-Rabin test is commonly used for this purpose.

3 There are also efficient algorithms to manufacture“certified” primes.

4 Are all large primes safe?

Factoring

Discussion1 To implement RSA we need to manufacture large primes.2 The Miller-Rabin test is commonly used for this purpose.3 There are also efficient algorithms to manufacture

“certified” primes.

4 Are all large primes safe?

Factoring

Discussion1 To implement RSA we need to manufacture large primes.2 The Miller-Rabin test is commonly used for this purpose.3 There are also efficient algorithms to manufacture

Square roots

Most integers are not perfect squares. Finding the square rootor identifying that it is not a perfect square is very easy. Yet inmodular arithmetic the situation is drastically different.

Half the positive integres mod a prime number p are quadraticresidues. While finding their square roots is not difficult it is a bittrickier than finding the square root of an integer.

Finding the square root of an integer mod p · q where p,q areprimes is dramatically different. Actually it is as difficult asfactoring. In other words, if there was a fast calculation of√

n mod p · q then we would have a fast factorization.

We shall start by learning how to find√

n mod p.

Square roots

n mod p.

Square roots

n mod p.

Square roots

n mod p.

√n mod p

1 Recall: GF (p) has primitve elements. Let α be a primitiveelement of GF (p).

2 n is a quadratic residue mod p if and only ifn = α2k mod p.

3 n is a quadratic residue mod p if and only ifn

p−12 mod p = 1.

4 Claim: If n is a quadratice residue mod p then we can findan integer β such that n2m+1β2s mod p = 1

n mod p = nm+1βs mod p6 We will not have to find a primitive element.

√n mod p

p−12 mod p = 1.

√n mod p

p−12 mod p = 1.

√n mod p

p−12 mod p = 1.

√n mod p

p−12 mod p = 1.

n mod p = nm+1βs mod p

6 We will not have to find a primitive element.

√n mod p

p−12 mod p = 1.

√n mod p

1 Calculate np−1

2 mod p = ±1 (use one of the powermodfunctions).

2 If np−1

2 mod p = −1 stop! n is not a quadratic residue.3 Let p − 1 = 2m(2k + 1).

4 Note: np−1

4 mod p = ±1. Repeat calculating np−1

2j until youget −1 or n2k+1 mod p = 1

5 If np−1

2j = −1 then find a non-quadratic residue β that isβ

p−12 mod p = −1 (easy, just try a few numbers).

6 np−1

2j βp−1

2 mod p = 1

7 Calculate: np−12j+1 β

p−14 mod p = ±1

8 Repeat the same process ubtil you reachn2k+1β2s mod p = 1

√n mod p

1 Calculate np−1

2 If np−1

2 mod p = −1 stop! n is not a quadratic residue.

3 Let p − 1 = 2m(2k + 1).

4 Note: np−1

5 If np−1

6 np−1

2j βp−1

2 mod p = 1

p−14 mod p = ±1

√n mod p

1 Calculate np−1

2 If np−1

4 Note: np−1

5 If np−1

6 np−1

2j βp−1

2 mod p = 1

p−14 mod p = ±1

√n mod p

1 Calculate np−1

2 If np−1

4 Note: np−1

5 If np−1

6 np−1

2j βp−1

2 mod p = 1

p−14 mod p = ±1

√n mod p

1 Calculate np−1

2 If np−1

4 Note: np−1

5 If np−1

6 np−1

2j βp−1

2 mod p = 1

p−14 mod p = ±1

√n mod p

1 Calculate np−1

2 If np−1

4 Note: np−1

5 If np−1

6 np−1

2j βp−1

2 mod p = 1

p−14 mod p = ±1

√n mod p

1 Calculate np−1

2 If np−1

4 Note: np−1

5 If np−1

6 np−1

2j βp−1

2 mod p = 1

p−14 mod p = ±1

√n mod p

1 Calculate np−1

2 If np−1

4 Note: np−1

5 If np−1

6 np−1

2j βp−1

2 mod p = 1

p−14 mod p = ±1

Examples

Example

1 p = 3 mod 4.2 This is a very easy case as

p−12 = 2k + 1, n2k+1 mod p = 1 so

√n mod p = nk+1.

3 Let p = 337639.71168819 mod 337639 = 1 (71 is a quadratic residue mod337639).71168820/2 mod 337639 = 234428.2344282 mod 337639 = 71.So√

71 mod 337639 = 234428.

Examples

Example1 p = 3 mod 4.

2 This is a very easy case asp−1

2 = 2k + 1, n2k+1 mod p = 1 so√

n mod p = nk+1.

71 mod 337639 = 234428.

Examples

Example1 p = 3 mod 4.2 This is a very easy case as

p−12 = 2k + 1, n2k+1 mod p = 1 so

√n mod p = nk+1.

71 mod 337639 = 234428.

Examples

p−12 = 2k + 1, n2k+1 mod p = 1 so

√n mod p = nk+1.

71 mod 337639 = 234428.

Examples

p−12 = 2k + 1, n2k+1 mod p = 1 so

√n mod p = nk+1.

3 Let p = 337639.

71168819 mod 337639 = 1 (71 is a quadratic residue mod337639).71168820/2 mod 337639 = 234428.2344282 mod 337639 = 71.So√

71 mod 337639 = 234428.

Examples

p−12 = 2k + 1, n2k+1 mod p = 1 so

√n mod p = nk+1.

3 Let p = 337639.71168819 mod 337639 = 1 (71 is a quadratic residue mod337639).

71168820/2 mod 337639 = 234428.2344282 mod 337639 = 71.So√

71 mod 337639 = 234428.

Examples

p−12 = 2k + 1, n2k+1 mod p = 1 so

√n mod p = nk+1.

3 Let p = 337639.71168819 mod 337639 = 1 (71 is a quadratic residue mod337639).71168820/2 mod 337639 = 234428.

2344282 mod 337639 = 71.So√

71 mod 337639 = 234428.

Examples

p−12 = 2k + 1, n2k+1 mod p = 1 so

√n mod p = nk+1.

71 mod 337639 = 234428.

1 p = 2701297, p − 1 = 24 · 33 · 132 · 37.

2 71(p−1)/2 = 2701296 so 71 is not a quadratic residue modp.

3 75 is a quadratic residue.4 75(p−1)/4 mod p = 2701296 = −1.5 71(p−1)/2 mod p = −1⇒ 75(p−1)/471(p−1)/2 mod p = 16 75(p−1)/871(p−1)/4 mod p = −1⇒

75(p−1)/8713(p−1)/4 mod p = 1.7 75(p−1)/16713(p−1)/8 = 75168331711012986 mod p = 18√

75 mod p = 75168332/2711012986/2 mod p = 23098919 Verify: 23098912 mod p = 75.

10 We can verify it in yet another way. 75 = 25 · 3. This meansthat√

3 mod p = 2309891/5. Indeed5−1 mod p = 1080519 and 1080519 · 2309891 mod p =1542497 and 15424972 mod p = 3.

1 p = 2701297, p − 1 = 24 · 33 · 132 · 37.2 71(p−1)/2 = 2701296 so 71 is not a quadratic residue mod

3 75 is a quadratic residue.4 75(p−1)/4 mod p = 2701296 = −1.5 71(p−1)/2 mod p = −1⇒ 75(p−1)/471(p−1)/2 mod p = 16 75(p−1)/871(p−1)/4 mod p = −1⇒

75(p−1)/8713(p−1)/4 mod p = 1.7 75(p−1)/16713(p−1)/8 = 75168331711012986 mod p = 18√

p.3 75 is a quadratic residue.

4 75(p−1)/4 mod p = 2701296 = −1.5 71(p−1)/2 mod p = −1⇒ 75(p−1)/471(p−1)/2 mod p = 16 75(p−1)/871(p−1)/4 mod p = −1⇒

75(p−1)/8713(p−1)/4 mod p = 1.7 75(p−1)/16713(p−1)/8 = 75168331711012986 mod p = 18√

p.3 75 is a quadratic residue.4 75(p−1)/4 mod p = 2701296 = −1.

5 71(p−1)/2 mod p = −1⇒ 75(p−1)/471(p−1)/2 mod p = 16 75(p−1)/871(p−1)/4 mod p = −1⇒

75(p−1)/8713(p−1)/4 mod p = 1.7 75(p−1)/16713(p−1)/8 = 75168331711012986 mod p = 18√

p.3 75 is a quadratic residue.4 75(p−1)/4 mod p = 2701296 = −1.5 71(p−1)/2 mod p = −1⇒ 75(p−1)/471(p−1)/2 mod p = 1

6 75(p−1)/871(p−1)/4 mod p = −1⇒75(p−1)/8713(p−1)/4 mod p = 1.

7 75(p−1)/16713(p−1)/8 = 75168331711012986 mod p = 18√

p.3 75 is a quadratic residue.4 75(p−1)/4 mod p = 2701296 = −1.5 71(p−1)/2 mod p = −1⇒ 75(p−1)/471(p−1)/2 mod p = 16 75(p−1)/871(p−1)/4 mod p = −1⇒

75(p−1)/8713(p−1)/4 mod p = 1.

7 75(p−1)/16713(p−1)/8 = 75168331711012986 mod p = 18√

75(p−1)/8713(p−1)/4 mod p = 1.7 75(p−1)/16713(p−1)/8 = 75168331711012986 mod p = 1

75(p−1)/8713(p−1)/4 mod p = 1.7 75(p−1)/16713(p−1)/8 = 75168331711012986 mod p = 18√

75 mod p = 75168332/2711012986/2 mod p = 2309891

9 Verify: 23098912 mod p = 75.10 We can verify it in yet another way. 75 = 25 · 3. This means

that√

75(p−1)/8713(p−1)/4 mod p = 1.7 75(p−1)/16713(p−1)/8 = 75168331711012986 mod p = 18√

An Application of Number Theory, the RSA Cryptosystemfaculty.washington.edu/moishe/hanoiex/Number...

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